Michael wrote:
> So pretty much the preorder needs to have a top or bottom element in order to define a well behaved identity for the monoid.

Hm. I'm not convinced that this is true in general, but my attempted counterexample foundered on associativity. The multiplication on the preorder below is not associative, since

$$c = Ic = (ab)c \ne a(bc) = ab = a.$$


a b
\ /
I
/ \
c d

|Iabcd
-+-----
I|Iabcd
a|aaIaa
b|bIbbb
c|ccccI
d|dddId